Find the limit of the sequence

... [itex]\{x_n\}[/itex] defined by [itex]x_1=1[/itex] and [itex]x_{n+1}=3+\sqrt{x_n}[/itex] for [itex]n\geq 2[/itex].

Here's what I did. We know for sure that a sequence as a limit if either it it increasing and has a superior bound or if it is decreasing and has an inferior bound. So let's suppose it satisfy either one of these condition and let's see what are the posible candidates for the limit. We supose that

[tex]\lim_{n\rightarrow \infty} x_n=x[/tex]

Now the limit of [itex]x_{n+1}[/itex] must [itex]x[/itex] too, because [itex]\{x_{n+1}\}[/itex] is a sub-sequence of [itex]\{x_n\}[/itex]. But we can find another expression for the limit of [itex]x_{n+1}[/itex], that is,